Question

If $\tanh x=\dfrac{4}{5}$, how do you find the values of the other hyperbolic functions at $x$ ?

Answer 1

We explain the function $\arctan \left( x \right)$. We express the inverse function of tan in the form of $\arctan \left( x \right)={{\tan }^{-1}}x$. It’s given that $\tanh x=\dfrac{4}{5}$. Thereafter we take all the other hyperbolic functions at $x$ of that angle to find the solution. We also use the representation of a right-angle triangle with height and base ratio being $\dfrac{4}{5}$ and the angle being $\theta$.

Complete step by step answer:
The hyperbolic functions are analogues of the ordinary trigonometric functions. All the usual relations are also used for the hyperbolic functions.It’s given that $\tanh x=\dfrac{4}{5}$. We can find the value of $\operatorname{sech}x$ from the relation of ${{\left( \operatorname{sech}x \right)}^{2}}=1+{{\left( \tanh x \right)}^{2}}$.
Putting the value, we get,
${{\left( \operatorname{sech}x \right)}^{2}}=1+{{\left( \dfrac{4}{5} \right)}^{2}}\\\ \Rightarrow{{\left( \operatorname{sech}x \right)}^{2}}=\dfrac{41}{25}$
Now taking square root we get

\Rightarrow\left( \operatorname{sech}x \right)=\dfrac{\sqrt{41}}{5}$$ Now we know the relation $$\cosh x=\dfrac{1}{\operatorname{sech}x}$$. Putting the value, we get $$\cosh x=\dfrac{1}{\operatorname{sech}x}\\\ \Rightarrow\cosh x =\dfrac{1}{\dfrac{\sqrt{41}}{5}}\\\ \Rightarrow\cosh x =\dfrac{5}{\sqrt{41}}$$ We know the sum of square law of ${{\left( \sinh x \right)}^{2}}+{{\left( \cosh x \right)}^{2}}=1$. Putting the value of $$\cosh x$$, we get ${{\left( \sinh x \right)}^{2}}=1-{{\left( \cosh x \right)}^{2}}\\\ \Rightarrow{{\left( \sinh x \right)}^{2}} =1-{{\left( \dfrac{5}{\sqrt{41}} \right)}^{2}}\\\ \Rightarrow{{\left( \sinh x \right)}^{2}} =\dfrac{16}{41}$ Taking square root, we get $\sinh x=\dfrac{4}{\sqrt{41}}$ We also have the relations $$\coth x=\dfrac{1}{\tanh x}$$ and $$\operatorname{csch}x=\dfrac{1}{\sinh x}$$. Putting the values, we get $$\coth x=\dfrac{1}{\tanh x}\\\ \Rightarrow\coth x =\dfrac{1}{\dfrac{4}{5}}\\\ \Rightarrow\coth x =\dfrac{5}{4}$$ And similarly, $$\operatorname{csch}x=\dfrac{1}{\sinh x}\\\ \Rightarrow\operatorname{csch}x =\dfrac{1}{\dfrac{4}{\sqrt{41}}}\\\ \Rightarrow\operatorname{csch}x =\dfrac{\sqrt{41}}{4}$$ **Hence, in this way we have found the all other hyperbolic functions.** **Note:** We can also apply the trigonometric triangle image form to get the value of other hyperbolic functions. In mathematics, hyperbolic functions are analogues of the ordinary trigonometric functions, but defined using the hyperbola rather than the circle. Just as the points form a circle with a unit radius, the points form the right half of the unit hyperbola.